We prove the statements simultaneously.
Note that by Lemma REF , \(\operatorname{Supp}_R(\operatorname{Ext}^i_R(\operatorname{Tr}M,N))\subseteq \operatorname{\mathcal {Z}}\)  for all \(i>0\) . It follows from Lemma REF  and assumption (iii) that \(\operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,N)=0\) .
By (REF .5), there exists the following exact sequence
\(\operatorname{Tor}_2^R(\Omega ^{-(n+1)}M,N)\rightarrow \operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,R)\otimes _RN \rightarrow \operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,N)\twoheadrightarrow \operatorname{Tor}_1^R(\Omega ^{-(n+1)}M,N).\qquad \mathrm {(\ref {l1}.1)}\) 
It follows from (REF .1) that \(\operatorname{Tor}_1^R(\Omega ^{-(n+1)}M,N)=0\) . As \(N\)  is Tor-rigid, we have
\(\operatorname{Tor}_j^R(\Omega ^{-(n+1)}M,N)=0 \text{ for all } j\ge 1.\qquad \mathrm {(\ref {l1}.2)}\) 
Again, by the exact sequence (REF .1), we get that \(\operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,R)\otimes _RN=0\)  which implies that \(\operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,R)=0\) . Note that by assumption (i) we have \(\operatorname{Supp}_R(\operatorname{Ext}^i_R(\operatorname{Tr}M,R))\subseteq \operatorname{\mathcal {Z}}\)  for all \(i\ge 1\) ; see [1]}. If \(\operatorname{\textnormal {grade}}_R(\operatorname{\mathcal {Z}},R)\ge n+1\) , then \(\operatorname{H}^{n}_{\operatorname{\mathcal {Z}}}(M)\cong \operatorname{Ext}^{n+1}_R(\operatorname{Tr}M,R)=0\)  by Lemma REF  and so the assertion (a) follows.
It follows from part (a) and (REF .7) that \(\Omega ^{-n}M\approx \Omega \Omega ^{-(n+1)}M\) . Therefore, by (REF .2), we get \(\operatorname{Tor}_j^R(\Omega ^{-n}M,N)=0\)  for all \(j>0\)  which proves part (b).
Now assume that \(n>0\) . By part (b) and (REF .1), we get the following isomorphism:
\(\operatorname{Ext}^{n}_R(\operatorname{Tr}M,R)\otimes _RN\cong \operatorname{Ext}^{n}_R(\operatorname{Tr}M,N).\qquad \mathrm {(\ref {l1}.3)}\) 
Now the assertion (c) follows from (REF .3) and Lemma REF .
